Hello, s_ingram!
Find the equations of the two circles which pass through the point (2,0)
and have both the $\displaystyle y$axis and the line $\displaystyle y\,=\, 1$ as tangents. Code:

 R
 +  P  * o *    y=1
+o:*
 * 2 : *
* : *
 :
* : *
Q o     o C *
* (x,y) *

* *
 * *
 * *
 * * *

The center of the circle is $\displaystyle C(x,y)$
It passes through the point $\displaystyle P(2,0)$
It is tangent to the $\displaystyle y$axis at $\displaystyle Q(0,y)$
It is tangent to $\displaystyle y = 1$ at $\displaystyle R(x,1)$
The radius is: .$\displaystyle CP \:=\: CQ \:=\: CR \:=\: r$
We have: .$\displaystyle x \:=\:r$
Since $\displaystyle CR \:=\:1y \:=\:r\:=\:x$, then: .$\displaystyle y \:=\:1x$
Since $\displaystyle CP \,=\,R$, we have: .$\displaystyle \sqrt{(x2)^2 + (y0)^2} \:=\:r$
. . which becomes: .$\displaystyle \sqrt{(x2)^2 + (1x)^2} \:=\:x$
Then: .$\displaystyle (x2)^2 + (1x)^2 \:=\:x^2$
. . which simplifies to: .$\displaystyle x^2  6x + 5 \:=\:0$
. . which factors: .$\displaystyle (x1)(x5) \:=\:0$
. . and has roots: .$\displaystyle x \:=\:1,5$
. . . . . . . . Then: .$\displaystyle y \:=\:0,\text{}4$
One circle has center $\displaystyle (1,0)$ and $\displaystyle r = 1\!:\quad \boxed{(x1)^2 + y^2 \:=\:1}$
The other has center $\displaystyle (5,\text{}4)$ and $\displaystyle r = 5\!:\quad \boxed{(x5)^2 + (y+4)^2 \:=\:25}$